Trying to use PHP variable in Javascript [duplicate]

Question

I am trying to bring a php variable into my javascript code to use for a countdown timer. Because of the other functions happening based on the timestamp php variable, I need to to be in php as well.

<?php  $access = '1443907640'; ?>
<script>
    $(function() {
        var access = <?php echo $access ?>;
        var note = $('.note'),

        ts = (new Date(access * 1000)).getTime() + 1 * 24 * 60 * 60 * 1000;


        $('.countdown').countdown({
        timestamp: ts,
        callback: function(days, hours, minutes, seconds) {

            var message = "";

            message += days + "<small class='opacity'>D</small>, ";
            message += hours + "<small class='opacity'>H</small>, ";
            message += minutes + "<small class='opacity'>M</small>, ";
            message += seconds + "<small class='opacity'>S</small>";

            note.html(message);

        }


        });
    });
</script>

Then I call call it with html here but doesn't work

<div class="note"></div>

Since you use Jquery, do you call it from within $(document).ready()? — some-non-descript-user
– some-non-descript-user, Commented Oct 3, 2015 at 21:39
Not seeing errors in console. I am mainly a php guy, not very good with js and jquery. I am only trying to use it so the countdown timer counts down by the second and isn't static — Dan Henry
– Dan Henry, Commented Oct 3, 2015 at 21:43
Just tried that solution, now seeing this in console ReferenceError: $ is not defined — Dan Henry
– Dan Henry, Commented Oct 3, 2015 at 21:45

Zakaria Acharki · Accepted Answer · 2015-10-03 22:00:13Z

1

I think the real problem in your code is in the end of this line :

var note = $('.note'),

Replace , by ; and it should work.

NOTE : line var access = <?php echo $access ?>; work fine in my test.

To solve the following error :

TypeError: $(...).countdown is not a function

You should add jquery.countdown script :

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery.countdown/2.1.0/jquery.countdown.js"></script>

Hope this helps.

edited Oct 3, 2015 at 22:00

answered Oct 3, 2015 at 21:48

Zakaria Acharki

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5 Comments

Dan Henry Over a year ago

Really? I tried that and am getting TypeError: $(...).countdown is not a function

Zakaria Acharki Over a year ago

You can see that i talk about a line and not the whole code... to solve the previous error see my updated answer.

Dan Henry Over a year ago

Ok did that, made the error disappear but still blank massblogprofits.com/get-launched/test.php

Zakaria Acharki Over a year ago

Ok, i cant see the class countdown in your code so please try to follow some tuto in jquery.countdown for a better implementation.

Dan Henry Over a year ago

Your solution plus forgetting to load my plugin, work now. Thanks!

Nibbels · Accepted Answer · 2015-10-03 21:50:18Z

0

Maybe consider to put an semicolon after your PHP-echo.

<?php echo $access; ?>

or the short-and-dirty version:

<?=$access?>

in line 4:

var access = <?=$access?>;

or

var access = '<?=$access?>'; //if $access would be typeof string

edited Oct 3, 2015 at 21:50

answered Oct 3, 2015 at 21:42

Nibbels

15610 bronze badges

4 Comments

some-non-descript-user Over a year ago

4th line in his code, isn't it?

Dan Henry Over a year ago

I have it as var access = "<?php echo $access; ?>";

Nibbels Over a year ago

yes! And you normally can omit the 'php' after those <? . I edited my post to make this clear.

Nibbels Over a year ago

I cannot comment the other Answer: If your page is blank, you have an PHP-Error somewhere.

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